An object with a mass of 2 kg is traveling in a circular path of a radius of 4 m. If the object's angular velocity changes from 3 Hz to 8 Hz in 2 s, what torque was applied to the object?

1 Answer
Aug 19, 2017

tau=503"Nm"

Explanation:

I will assume that you meant that the frequency changes from 3"Hz" to 8"Hz" (which in turn will affect the angular velocity).

Torque can be expressed as the rate of change of angular momentum.

(1)" "color(darkblue)(tau=(dL)/(dt))

Angular momentum can be described by the equation:

L=Iomega

where I is the moment of inertia and omega is the angular velocity

Substituting this into equation (1), we have:

(2)" "tau=(d(Iomega))/(dt)

The angular velocity an be expressed by the equation:

omega=2pif

Substituting this into equation 3, we get:

(3)" "tau=(d(I*2pif))/(dt)

Because we are given that the frequency of the motion is changing we can move I and 2pi outside of the differential as constants and rewrite equation (3) as:

(4)" "tau=2piI*(df)/(dt)

Finally, for a (~~point) mass traveling a circular path about a center axis, the moment of inertia is given by I=mr^2.

Additionally, we will use the average rate of change, meaning that

(df)/(dt)=(Deltaf)/(Deltat)=(f_f-f_i)/(t_f-t_i).

Substituting both of the above into equation (4), we get:

(5)" "color(darkblue)(tau=2pimr^2*(f_f-f_i)/(t_f-t_i)

We are given the following information:

  • |->m=2"kg"
  • |->r=4"m"
  • |->t=2"s"
  • |->f_i=3"s"^-1
  • |->f_f=8"s"^-1

Substituting these values into equation (5), we get:

tau=2pi(2"kg")(4"m")^2*(8"s"^-1-3"s"^-1)/(2"s"-0"s")

=>=64pi"kgm"^2*5/2s^-2

=>=160pi"Nm"

=>502.655"Nm"

:.tau~~503"Nm"